University Chemistry
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Properties of Gases
• Gases have four main characteristics
compared with solids and liquids:
• Gases take the volume and shape of their containers.
• Mix completely (homogeneously) with any other gas.
• Compressible: their volume decrease when pressure
applied.
• Exert pressure on its surroundings.
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Pressure of Gases and its Units
Pressure is defined as the force applied per unit area.
Standard atmospheric pressure: the
pressure that supports a column of
mercury exactly 760 mm high at 0oC at
sea level.
A barometer is used to
measure the pressure as shown.
Common Units of Pressure
SI units of pressure is kg/m. s2
Pascal, Pa = Newton/meter2 = N/m2 (N = kg m/s2)
Standard Atmospheric Pressure = 101.3 kPa
=1 atmosphere (atm)
= 760 torr
= 760 mm Hg
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Gas Laws
Boyle’s Law:
(the relationship between volume and pressure):
“The volume of a sample of gas is inversely proportional to its
pressure, if temperature remains constant”
Pressure x Volume = Constant (T = constant)
P1 and V1 are initial conditions
P2 and V2 are final conditions
P1V1 = P2V2 (T = constant)
Example
A cylinder of oxygen has V = 50.0 L and P = 10 atm at 20oC. What will
be the volume of the gas at atmospheric pressure (1 atm) and 20oC?
According to Boyle’s law
P1V1 = P2V2
P1= 10 atm, V1 = 50.0 L, P2= 1 atm , V2 = ??
(10 atm x 50 L) = (1 atm x V2), and V2 = 500 L
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Gas Laws
Charles’s Law:
(the relationship between volume and temperature):
“The volume of a fixed amount of gas is directly proportional to the
absolute temperature of the gas at constant pressure.”
Volume = Temperature x Constant
(P = constant)
V1 V2
=
T1 T2
Example
A sample of gas at 1 atm pressure and 25oC has V = 1 L, The gas is
cooled to -196 oC. What is the new volume?
According to Charles’ law
V1/T1 = V2/T2
V1= 1.0 L, T1 = 25oC = 25+ 273.15 = 298.15 K,
V2= ?? L, T2 = -196oC = -196 + 273.15 = 76.15 K
V1/T1 = V2/T2
(1.0 L / 298.15 K) = (V2 L / 76.15 K)
V2= 0.26 L
(temperature decreased, volume decreased)
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Gas Laws
Avogadro’s Law:
(the relationship between volume and amount):
“The volume of a gas is directly proportional to the number of moles (n)
of gas at constant temperature and pressure.”
Volume = # moles x Constant
V1 V2
=
n1 n 2
Example
In the following reaction: 3 H2 (g) + N2 (g) → 2 NH3 (g)
Before the reaction begins, the reactants occupy a volume of 1.32 L. If a constant
pressure is maintained in the reaction, what is the final volume occupied by the
product after the completion of the reaction (assume all reactants are consumed)?
According to Avogadro’s law
V1/n1 = V2/n2
V1= 1.320 L, n1 = 3 mole (H2) + 1 mole (N2) = 4 moles,
V2= ?? L, n2 = 2 moles (NH3)
V1/n1 = V2/n2
(1.32 L / 4 moles K) = (V2 L / 2 moles)
V2= 0.660 L
(number of moles decreased, volume decreased)
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Ideal Gas Law
Boyle’s Law:
V α 1/P (T = constant)
Charles’s Law:
V α T (P = constant)
Avogadro’s Law: V α n
T and P are constant
Therefore: V α nT/P
or
PV α nT
So, the Ideal Gas Law: PV = nRT
Where: R = proportionality constant called the universal
gas constant and it is equal = 0.08206 L atm K-1 mol-1,
or 8.3145 J mol-1. K-1.
P = pressure in atm
V = volume in Litres
n = moles
T = temperature in Kelvin
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Example
What is the lung capacity (volume) of an average adult lung, if
the number of moles of the air in the lungs is 0.15 moles?
Assume that the person is at 1.00 atm pressure and has a
normal body temperature of 37oC.
Answer
According to the ideal gas law
PV = nRT
P = 1.00 atm, V= ??, n = 0.15 moles, R = 0.08206 L atm K-1
mol-1,
T = 37 oC = 37 + 273.15 = 310.15 K
V = nRT/P
= (0.15 mole x 0.08206 L atm K-1 mol-1 x 310.15 K)/1.00 atm
The lung capacity V = 3.8 L
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Standard Temperature and Pressue “STP”
The STP are considered when the pressure is equal to 1 atm
and the temperature = 0 oC (273 K).
Example
What is the molar volume (volume of 1 mole) of an ideal gas at STP?
According to the ideal gas law: PV = nRT
• V = (nRT)/P = (1 mole x 0.08206 L atm K-1 mol-1 x 298.15 K)/ (1.00 atm)
• Molar volume Vm = 22.4 L
• The molar volume of ANY ideal gas is 22.4 liters at STP.
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Dalton’s Law of Partial Pressure:
For a mixture of gases in a container, the overall pressure is the sum of
all the partial pressures of the individual components.
PTotal = P1 + P2 + P3 + . . .
Example
A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure
of 278 kPa. If the partial pressures of the oxygen and the hydrogen are 112
kPa and 101 kPa respectively, what would be the partial pressure exerted
by the nitrogen.
According to Dalton’s law of partial pressure
Ptotal = P1 + P2 + . . . Pn
278 kPa = 112 kPa + 101 kPa + Pnitrogen
Pnitrogen = 278 kPa - (112 kPa + 101 kPa)
Pnitrogen = 65 kPa
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Mole Fraction (X)
• If a mixture contains two components, A
and B, then:
moles A
XA =
moles A + moles B
XB =
moles B
moles A + moles B
XA + XB = 1
Example
A gaseous mixture contains 0.38 moles of nitrogen gas and 0.45 moles
of oxygen gas. Determine the mole fractions of oxygen and nitrogen.
Total number of moles = 0.38 mol nitrogen + 0.45 mole oxygen = 0.83 moles
XOxygen = (moles of oxygen / total number of moles)
= (0.45 mole/ 0.83 moles) = 0.54
Xnitrogen = (moles of nitrogen / total number of moles)
= (0.38 mole/ 0.83 moles) = 0.46
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Check: XOxygen + Xnitrogen = 0.54 + 0.46 =1.00
Determining Partial Pressure from Mole Fraction
XA =
n A RT
PA =
V
PA
Ptotal
moles A
moles of all species in sample
Ptotal
n total RT
=
V
 n A RT 

 n
V
 = A = XA
=
 n total RT  n total


 V 
PA = Ptotal X A
Example
The mole fraction of nitrogen in air is 0.7808. Calculate the partial
pressure of N2 in air when the atmospheric pressure is 760 torr.
Use the partial pressure formula:
PN 2 = PtotalX N 2 = (760 torr)(0.7808) = 593 torr
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Graham’s Law of diffusion and effusion of gases
Gas Diffusion is the gradual mixing of
molecules of different gases. A good
example is the diffusion of oxygen in the
lungs through the alveoli.
Gas Effusion is the escape of a gas from
its container through a small hole.
Graham’s Law stated that "The relative rates at which two gases
under identical conditions of temperature and pressure will diffuse vary
inversely as the square roots of the molecular masses of the gases."
Rate A
MWB
=
Rate B
MWB
Example
Calculate the relative rates of effusion of H2(g) and He(g)
Rate H 2
Rate He
=
MWHe
4
=
= 2 = 1.41
MWH 2
2
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University Chemistry
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